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I need to find the $V$ by triple integral.
the limits from up is (1) - ecliptic cone.
and from dwon - (2) - elepsoide.

$$(1) \ \ \ \ z=-\sqrt{3x^2+5y^2}$$ $$(2) \ \ \ \ {3 \over 10}x^2+5y^2+{z^2 \over 3}=8$$

enter image description here

I try to use:
$x=5r\cos\theta$
$y=3r\sin\theta$
$z=z$

but I'm not sure it's good coordinate and I can't find the limit.

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  • $\begingroup$ I don't think these should be called elliptic coordinates. If this is what you meant by ecliptic coordinates. This is rather (almost) polar coordinates for $x,y$. $\endgroup$ – Julien Mar 15 '13 at 13:07
  • $\begingroup$ sorry, you right $\endgroup$ – user1816377 Mar 15 '13 at 13:09
  • $\begingroup$ It's still not clear to me what the volume looks like - can you be a little more descriptive? $\endgroup$ – Ron Gordon Mar 15 '13 at 13:20
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    $\begingroup$ So you want to find the volume outside the cone but inside the ellipsoid? $\endgroup$ – Ron Gordon Mar 15 '13 at 13:53
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    $\begingroup$ @user1816377: I added a plot. If you don't like it remove it. :-) $\endgroup$ – mrs Mar 15 '13 at 15:01
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This one is obviously a little messy, but there is a simple methodology by which the volume may be reduced to a single integral. The idea is to compute the cross sectional area $A(z)$ for each value of $z$ in the volume, and then integrate over $z$ to get the volume.

First, we needs the bounds in $z$. This is done by observing that the relevant $z$ vertex of the ellipsoid is at $z=-2 \sqrt{6}$. In between, it should be clear that each cross section is an intersection of two ellipses centered at the origin. The first has equation

$$\frac{x^2}{z^2/3} + \frac{y^2}{z^2/5} = 1$$

the other has equation

$$\frac{x^2}{\frac{80}{3} \left(1-\frac{z^2}{24}\right)} + \frac{y^2}{\frac{8}{5} \left(1-\frac{z^2}{24}\right)}=1$$

There are three different scenarios: one in which the first ellipse is entirely within the second, one in which the second is entirely within the first, and one in which they stick out of each other. The intervals in $z$ where these different scenarios occur may be determined by equating the vertices in $x$ and $y$, respectively, in the first and second ellipses. The result is three intervals in $z$:

  • I: $z \in \left[-2 \sqrt{6},-\sqrt{\frac{240}{13}}\right]$
  • II: $z \in \left[-\sqrt{\frac{240}{13}},-\sqrt{6}\right]$
  • III: $z \in \left[-\sqrt{6},0\right]$

In interval I, the area is simply the area of the ellipsoid cross section:

$$A(z) = \frac{8 \sqrt{6}}{3} \pi \left(1-\frac{z^2}{24}\right)$$

(Note that I used the formula $A=\pi a b$ as the area of the ellipse $(x^2/a^2)+(y^2/b^2)=1$.)

In interval III, the area is simply the area of the conical cross section:

$$A(z) = \frac{\pi}{\sqrt{15}} z^2$$

That leaves interval II, in which there is an intersection:

ellipses

The cross sectional area is broken up into 2 pieces: one bounded by the first ellipse at the sides, and another bounded by the second ellipse in the center. The area is then

$$4 \int_{x_0(z)}^{z/\sqrt{3}} dx \: \frac{1}{\sqrt{5}}\sqrt{z^2-3 y^2} + 2 \int_{-x_0(z)}^{x_0(z)} dx \: \sqrt{\frac{8}{5}} \sqrt{1-\frac{z^2}{24}-3 \frac{x^2}{80}}$$

where

$$x_0(z)=\frac{\sqrt{z (z+3)-24}}{3 \sqrt{\frac{1}{z}-\frac{1}{10}}}$$

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  • $\begingroup$ To make the computations easier I thought of making a change of variables so that the ellipsoid becomes a sphere, and modify the elliptical cone accordingly. I am not sure if this is of much help though. $\endgroup$ – Américo Tavares Apr 22 '13 at 23:34
  • $\begingroup$ @AméricoTavares: worth a try; this problem is a bit sadistic to be just a homework problem IMO. $\endgroup$ – Ron Gordon Apr 22 '13 at 23:37
  • $\begingroup$ That's right! BTW I gave a look at the solution you posted in your blog to the "3 cylinders" challenge. $\endgroup$ – Américo Tavares Apr 22 '13 at 23:41
  • $\begingroup$ @AméricoTavares: Oh, terrific, glad someone's looking at it. Was the solution understandable? $\endgroup$ – Ron Gordon Apr 22 '13 at 23:42
  • $\begingroup$ Yes, but I didn't check every detail. It is amazing how fast you posted it. I had the idea of just computing the volume in the 1st octant and multiply the result by 8, but I am very slow. $\endgroup$ – Américo Tavares Apr 22 '13 at 23:47

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